# [SOLVED]Velleman problem 14(a) section 7.3

#### issacnewton

##### Member
Hi
I have to prove that $$^{\mathbb{R}}\mathbb{R}\;\sim\;\mathcal{P}(\mathbb{R})$$.
My attempt is here. $$\mathbb{R}\;\sim\;\mathbb{R} \Rightarrow \mathbb{R}\;\precsim\;\mathbb{R}$$. Since
$$\{0,1\}\subseteq \mathbb{R}\Rightarrow \{0,1\}\;\precsim\;\mathbb{R}$$ . I am going to make use of the rule which I have proven.
if $$A\neq\varnothing$$ and $$A\;\precsim\; B$$ and $$C\;\precsim\; D$$ then $$^{A}C\;\precsim\; ^{B}D$$. So we get
$$^{\mathbb{R}}\{0,1\}\;\precsim\; ^{\mathbb{R}}\mathbb{R}$$. Since $$\mathcal{P}(\mathbb{R})\;\sim\; ^{\mathbb{R}}\{0,1\}$$, it
follows that $$\mathcal{P}(\mathbb{R})\;\precsim\; ^{\mathbb{R}}\{0,1\}$$. So using transitivity of $$\precsim$$ we get
$$\mathcal{P}(\mathbb{R})\;\precsim\;^{\mathbb{R}} \mathbb{R} \cdots (E1)$$. Now

$^{\mathbb{R}}\mathcal{P}(\mathbb{R})\;\sim\;^{ \mathbb{R}}( ^{\mathbb{R}}\{0,1\})\;\sim\;^{(\mathbb{R}\times \mathbb{R})}\{0,1\}$

But since $$\mathbb{R}\times\mathbb{R}\;\sim\; \mathbb{R}$$ we have

$^{(\mathbb{R}\times\mathbb{R})}\{0,1\}\;\sim\; ^{\mathbb{R}}\{0,1\}\;\sim\; \mathcal{P}(\mathbb{R})$

which implies, due to the transitivity of $$\sim$$

$^{\mathbb{R}}\mathcal{P}(\mathbb{R})\;\sim\; \mathcal{P}(\mathbb{R})$

$\Rightarrow ^{\mathbb{R}}\mathcal{P}(\mathbb{R})\;\precsim\; \mathcal{P}(\mathbb{R})$

Now

$\mathbb{R}\;\precsim\;\mathcal{P}(\mathbb{R}); \; \mathbb{R}\;\precsim\; \mathbb{R}$

since $$\mathbb{R}\neq \varnothing$$ , we get

$^{\mathbb{R}}\mathbb{R}\;\precsim\; ^{\mathbb{R}}\mathcal{P}(\mathbb{R})\;\precsim\; \mathcal{P}(\mathbb{R}) \cdots (E2)$

Using E1 and E2 , it follows from Cantor-Schroder-Bernstein theorem, that

$^{\mathbb{R}}\mathbb{R}\;\sim\;\mathcal{P}(\mathbb{R})$

does it seem ok ? I have already proved all the identities I am using here...

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Yes, this also seems fine. Some transitions can be shortened: for example, from $A\precsim B$ and $B\sim C$ you can directly conclude $A\precsim C$ without stating that $B\precsim C$.